Problem: $\int \dfrac{x}{\sqrt{16-x^2}}\,dx\,= $ $+~C$
Explanation: Notice that we can rewrite the integral as $ \int \dfrac{1}{\sqrt{16-x^2}}\cdot \,x\, dx\,$. If we let $ {u=16-x^2}$, then ${du=-2x \, dx}$ and $x\, dx=-\dfrac{du}{2}}$. Substituting gives us: $\begin{aligned} \int \dfrac{1}{\sqrt{{16-x^2}}}\cdot \,x\, dx}\,&= \int \dfrac{1}{\sqrt{ u}}\,\left( -\dfrac{du}{2}}\right)\\\\\\ &= \int \dfrac{1}{\sqrt{ u}}\cdot \left(-\dfrac{1}{2}\right)\,du\\\\\\ &=-\dfrac{1}{2} \int \dfrac{1}{\sqrt{ u}}\, du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= -\dfrac{1}{2} \int \dfrac{1}{\sqrt{ u}}\, du\\\\\\ &=-\dfrac{1}{2}\cdot 2\sqrt u+C\\\\\\ &=-\sqrt{u}+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = − u √ + C = − 16 − x 2 − − − − − − √ + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=-\sqrt{u}+C\\\\\\\ &=-\sqrt{16-x^2}+C\end{aligned} The answer: $\int \dfrac{x}{\sqrt{16-x^2}}\,dx\,= -\sqrt{16-x^2}+C$